Chapter 2 Mechanics of Materials Tensile stress (+) Compressive stress (-) Normal stress = At a point: Example: Estimate the normal stress on a shin bone (脛骨)

Shear stress (切應力) =  = F tangential to the area / A At a point,

Shear strain (?) = deformation under shear stress = Normal strain (正應變)  = fractional change of length= l x Shear strain (?) = deformation under shear stress = x F fixed

Hooke's law: In elastic region,   , or /  = E Stress-strain curve   o Yield pt. Work hardening break Elastic deformation Hooke's law: In elastic region,   , or /  = E E is a constant, named as Young’s modulus or modulus of elasticity Similarly, in elastic region, / = G, where G is a constant, named as shear modulus or modulus of rigidity.

Exercise set 2 (Problem 3) Find the total extension of the bar. 15mm W 5mm 1.2m 0.6m o dx Width of a cross-sectional element at x: Stress in this element : Strain of this element: The extension of this element : The total extension of the whole bar is : = 2.13 x 10-4 m

Bulk modulus 

Poisson's ratio : For a homogeneous isotropic material x d normal strain : lateral strain : Poisson's ratio : value of  : 0.2 - 0.5

Double index notation for stress and strain 1st index: surface, 2nd index: force For normal stress components : x  xx, y  yy , z  zz, x  xx y z zx zy yz xz xy yx

Joint effect of three normal stress components y z

Symmetry of shear stress components Take moment about the z axis, total torque = 0, (xy yz) x = (yx x z) y, hence, xy = yx . Similarly, yz = zy and xz = zx z y x xy yx x y z

Original shear strain is “simple” strain = x dy rot 2 There is no real deformation during pure rotation, but “simple” strain  0. Define pure rotation angle rot and pure shear strain, such that the angular displacements of the two surfaces are: 1= rot+ def and 2= rot- def . Hence, rot = (1+ 2)/2 and def = (1- 2)/2 def x 2 = - Example: 1 = 0 and 2 = - , so def = (0+)/2 = /2 and rot= (0-)/2 = -/2 Pure shear strain is /2

For hydrostatic pressure Example: Show that Proof: For hydrostatic pressure l xx = yy = zz = , hence 3 = xx+yy+zz = (1-2v)(xx+yy+zz)/E xx =yy =zz = -p (compressive stress)

Example : Show that nn = /2 2019/7/17 Example : Show that nn = /2 Point C moves further along x- and y-direction by distances of AD(/2) and AD(/2) respectively. nn = [(AD  /2)2 + (AD  /2)2]1/2 / [(AD)2 + (AD)2]1/2 = /2 True shear strain: yx = /2 Therefore, the normal component of strain is equal to the shear component of strain: nn = yx and nn = /2 A C’ C D D’

Consider equilibrium along n-direction: Example : Show that nn = nn/(2G) Consider equilibrium along n-direction:  yx (lW) sin 45o x2 = 2 (l cos 45o) W nn l n yx xy 2 l cos 45o Therefore yx = nn From definition :  = xy /G = nn /G = 2 nn

Example : Show xx = xx/E -  yy/E- v zz/E Set xx =  nn = - yy,  zz = 0, xx = nn nn = (1+)  nn /E =  nn /2G (previous example) nn -nn

xx = (xx- v yy- v zz) /E 0 = [xx- 0 – 0.3(- 3107)]/60109 12kN x Ex. 12 kN forces are applied to the top & bottom of a cube (20 mm edges), E = 60 GPa,  = 0.3. Find (i) the force exerted by the walls, (ii) yy xx = 0, yy = 0 and zz= -12103 N/(2010-3 m)2 = 3107 Pa xx = (xx- v yy- v zz) /E 0 = [xx- 0 – 0.3(- 3107)]/60109  xx = -9106 Pa (compressive) Force = Axx = (2010-3 m)2(-9106 Pa) = -3.6 103 N (ii) yy = (yy- v zz- v xx) /E = [0 – 0.3(- 3107) – 0.3(- 9106)]/60109 = 1.9510-4

Elastic Strain Energy The energy stored in a small volume:  Energy density in the material : e=extension x

Similarly for shear strain :